Simplifying the Trigonometric Expression sin 3x cos x cos 3x sin x
Introduction
Trigonometric expressions often look intimidating, especially when they combine multiple angles like sin 3x cos x cos 3x sin x. Mastering the techniques to simplify such products not only sharpens your algebraic skills but also deepens your understanding of fundamental trigonometric identities. In this article we will walk through a clear, step‑by‑step process to reduce sin 3x cos x cos 3x sin x into a compact form, explore the scientific reasoning behind each transformation, and answer common questions that arise when working with these types of problems Worth keeping that in mind..
Steps to Simplify sin 3x cos x cos 3x sin x
Step 1: Recognize the structure
The given expression is a product of four trigonometric terms:
[ \sin 3x ;\cos x ;\cos 3x ;\sin x ]
Notice that the angles appear in pairs: (3x) and (x). This symmetry suggests that we can apply product‑to‑sum identities or double‑angle formulas to combine them efficiently.
Step 2: Pair the terms
We can group the terms as ((\sin 3x \cos x)) and ((\cos 3x \sin x)). Each pair resembles the pattern (\sin A \cos B). The product‑to‑sum identity for this pattern is:
[ \sin A \cos B = \frac{1}{2}\big[\sin(A+B) + \sin(A-B)\big] ]
Applying this to the first pair ((A = 3x, B = x)):
[ \sin 3x \cos x = \frac{1}{2}\big[\sin(3x + x) + \sin(3x - x)\big] = \frac{1}{2}\big[\sin 4x + \sin 2x\big] ]
Similarly, for the second pair ((A = 3x, B = x) but swapped order, note that (\cos 3x \sin x = \sin x \cos 3x)):
[ \cos 3x \sin x = \frac{1}{2}\big[\sin(x + 3x) + \sin(x - 3x)\big] = \frac{1}{2}\big[\sin 4x + \sin(-2x)\big] ]
Since (\sin(-\theta) = -\sin\theta), this simplifies to:
[ \cos 3x \sin x = \frac{1}{2}\big[\sin 4x - \sin 2x\big] ]
Step 3: Multiply the two results
Now we have:
[ \sin 3x \cos x \cos 3x \sin x = \left(\frac{1}{2}[\sin 4x + \sin 2x]\right) \times \left(\frac{1}{2}[\sin 4x - \sin 2x]\right) ]
Factor out the (\frac{1}{4}):
[ = \frac{1}{4}\big(\sin 4x + \sin 2x\big)\big(\sin 4x - \sin 2x\big) ]
This is a classic difference of squares:
[ (a+b)(a-b) = a^{2} - b^{2} ]
Let (a = \sin 4x) and (b = \sin 2x). Then:
[ = \frac{1}{4}\big(\sin^{2}4x - \sin^{2}2x\big) ]
Step 4: Apply the power‑reduction identity
To simplify (\sin^{2}\theta) we use the power‑reduction formula:
[ \sin^{2}\theta = \frac{1 - \cos 2\theta}{2} ]
Applying this to both terms:
[ \sin^{2}4x = \frac{1 - \cos 8x}{2}, \qquad \sin^{2}2x = \frac{1 - \cos 4x}{2} ]
Plug them back:
[ \frac{1}{4}\left[\frac{1 - \cos 8x}{2} - \frac{1 - \cos 4x}{2}\right] = \frac{1}{4}\cdot\frac{1}{2}\big[(1 - \cos 8x) - (1 - \cos 4x)\big] ]
Simplify the numerator:
[ = \frac{1}{8}\big[1 - \cos 8x - 1 + \cos 4x\big] = \frac{1}{8}\big[\cos 4x - \cos 8x\big] ]
Step 5: Use the sum‑to‑product identity
The expression (\cos 4x - \cos 8x) can be transformed with the identity:
[ \cos A - \cos B = -2\sin!\left(\frac{A+B}{2}\right)\sin!\left(\frac{A-B}{2}\right) ]
Let (A = 4x) and (B = 8x):
[ \cos 4x - \cos 8x = -2\sin!\left(\frac{4x + 8x}{2}\right)\sin!\left(\frac{4x - 8x}{2}\right) = -2\sin 6x \sin(-2x) ]
Since (\sin(-\theta) = -\sin\theta):
[ = -2\sin 6x \cdot (-\sin 2x) = 2\sin 6x \sin 2x ]
Now substitute back:
[ \frac{1}{8}\big[\cos 4x - \cos 8x\big] = \frac{1}{8}\big[2\sin 6x \sin 2x\big] = \frac{1}{4}\sin 6x \sin 2x ]
Step 6: Final compact form
Thus, the original product simplifies to:
[ \boxed{\displaystyle \sin 3x \cos x \cos 3x \sin x = \frac{1}{4}\sin 6x \sin 2x} ]
If desired, we can apply the product‑to‑sum identity one more time to express the result as a sum of sines:
[ \frac{1}{4}\sin 6x \sin 2x = \frac{1}{8}\big[\cos(6x-2x) - \cos(6x+2x)\big] = \frac{1}{8}\big[\cos 4x - \cos 8x\big] ]
Both forms are equivalent; the choice depends on the context of the problem.
Scientific Explanation
The simplification process relies on three core trigonometric identities:
- Product‑to‑Sum: (\sin A \cos B = \frac{1}{2}[\sin(A+B
